on the rational polytopes with chvatal rank...
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On the Rational Polytopes with Chvatal Rank 1
Gerard Cornuejols∗ Dabeen Lee† Yanjun Li‡
November 30, 2016
Abstract
In this paper, we study the following problem: given a polytope P with Chvatal rank 1,does P contain an integer point? Boyd and Pulleyblank observed that this problem is in thecomplexity class NP ∩ co-NP, an indication that it is probably not NP-complete. We solvethis problem in polynomial time for polytopes arising from the satisfiability problem of a givenformula with at least three literals per clause, for simplices whose integer hull can be obtainedby adding at most a constant number of Chvatal inequalities, and for rounded polytopes. Weprove that any closed convex set whose Chvatal closure is empty has an integer width of atmost n, and we give an example showing that this bound is tight within an additive constantof 1. The promise that a polytope has Chvatal rank 1 seems hard to verify though. We provethat deciding emptiness of the Chvatal closure of a given rational polytope P is NP-complete,even when P is contained in the unit hypercube or is a rational simplex, and even when P doesnot contain any integer point. This has two implications: (i) It is NP-hard to decide whethera given rational polytope P has Chvatal rank 1, even when P is contained in the unit cube oris a rational simplex; (ii) The optimization and separation problems over the Chvatal closureof a given rational polytope contained in the unit hypercube or of a given rational simplex areNP-hard. These results improve earlier complexity results of Cornuejols and Li and Eisenbrand.Finally, we prove that, for any positive integer k, it is NP-hard to decide whether adding atmost k Chvatal inequalities is sufficient to describe the integer hull of a given rational polytope.
1 Introduction
Let P ⊂ Rn be a rational polyhedron, and let PI denote its integer hull, namely PI := conv(P ∩Zn),the convex hull of all the integer points in P . If an inequality cx ≤ d is valid for P for some c ∈ Zn,then cx ≤ ⌊d⌋ is a valid inequality for PI . Chvatal [6] introduced the following beautiful notion ofclosure.
P ′ :=⋂
c∈Zn
{x ∈ Rn : cx ≤ ⌊max{cx : x ∈ P}⌋}
It follows from the definition that PI ⊆ P ′ ⊆ P . The set P ′ is called the Chvatal closure of P .Although P ′ is defined as the intersection of infinitely many half-spaces, P ′ turns out to be arational polytope when P is a rational polytope [6]. Schrijver [30] later extended this result torational polyhedra. Bockmayr and Eisenbrand [3] proved that the number of inequalities neededto describe P ′ is polynomially bounded in fixed dimension.
∗Tepper School of Business, Carnegie Mellon University, Pittsburgh, PA 15206, USA, [email protected]†Tepper School of Business, Carnegie Mellon University, Pittsburgh, PA 15206, USA, [email protected]‡Krannert School of Management, Purdue University, West Lafayette, IN 47906, USA, [email protected]
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The problem of deciding whether a rational polytope contains an integer point is NP-complete [4].The main motivation of this paper is the following question: for a rational polytope with Chvatalrank 1, can the integer feasibility problem be solved in polynomial time? Boyd and Pulleyblank [5]observed that this problem belongs to the complexity class NP∩ co-NP. This is an indication thatthis problem is not NP-complete (unless NP = co-NP). In Section 2 we give polynomial algorithmsfor several special cases of this problem.
Although researchers have investigated the Chvatal closure of polytopes associated with severalcombinatorial optimization problems, such as the fractional stable set [18] and matching poly-topes [15], a constructive characterization of the Chvatal closure of an arbitrary polytope is notwell studied. Giles and Pulleyblank [19] showed that, for every rational polyhedron, there exists atotally dual integral (TDI) system Ax ≤ b where A is an integral matrix. The Chvatal closure canbe obtained by just rounding the right hand side of the integral TDI system [30]. However, findingan integral TDI system is not an easy task, so this approach is not efficient in characterizing theChvatal closure.
In Section 3, we prove that any closed convex set in Rn whose Chvatal closure is empty hasan integer width of at most n, and we give an example showing that this bound is tight withinan additive constant of 1. We remark that the upper bound on the width implies the existenceof a deterministic 2O(n)nn Lenstra-type algorithm for the integer feasibility problem over a givenrational polyhedron with Chvatal rank 1. On the other hand, the lower bound indicates that wecannot improve this time complexity if we use a Lenstra-type procedure.
Letchford, Pokutta and Schulz [26] asked whether the separation problem for the Chvatalclosure remains NP-hard for rational polytopes contained in the unit hypercube. We prove this inSection 4.1. In fact, we prove that deciding emptiness of the Chvatal closure of a rational polytopeP contained in the unit hypercube [0, 1]n is NP-complete, even when P contains no integer point.This result implies that optimizing over the Chvatal closure of a rational polytope contained in theunit hypercube is NP-hard by the equivalence between optimization and separation [20].
We also study the same computational complexity question for a full-dimensional rational sim-plex in Rn, which is the convex hull of n + 1 affinely independent points. Sebo [32] observed thatinteger programming over rational simplices is NP-complete. In Section 4.2, we show that theproblem of deciding emptiness of the Chvatal closure of a given rational simplex is NP-complete.Strong NP-completeness is still an open problem.
In Section 5, we investigate the problem of deciding whether a given rational polytope hasChvatal rank 1. Cornuejols and Li [9] showed that the problem is NP-hard. We prove that theproblem remains NP-hard even when the polytope is contained in the unit hypercube or is a simplex.This result answers an open question raised in [9].
In Section 6, we study the problem of deciding whether we can obtain the integer hull ofa polyhedron P by adding at most k Chvatal inequalities to the description of P for a givennonnegative integer k. In the case of k = 0, we know that it is strongly co-NP-complete to decidewhether a given polyhedron is integral [14, 28]. In the case of k ≥ 1, we prove that deciding whetherthe integer hull of a polyhedron P can be obtained by adding at most k Chvatal inequalities isNP-hard. Previously, Mahajan and Ralphs [27] showed that deciding whether a given rationalpolytope is contained in a strip, where a strip is a set {x ∈ Rn : π0 < x < π0 + 1} for some(π, π0) ∈ Zn+1, is NP-complete. We borrow some of their ideas to prove our NP-hardness result.
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2 Polynomially solvable cases and open problems
In this section, we consider the problem of deciding whether a rational polyhedron P contains aninteger point under the promise that P has Chvatal rank 1. This promise on the input P verylikely modifies the computational complexity of the integer feasibility problem.
Theorem 1 (Boyd and Pulleyblank [5]). Let P ⊂ Rn be a rational polyhedron with Chvatal rank 1.The problem of deciding whether P contains an integer point is in the complexity class NP∩co-NP.
The problems in NP ∩ co-NP are probably not NP-complete (since otherwise NP = co-NP), so wehave the following question:
Open Question 1. Let P ⊂ Rn be a rational polyhedron with Chvatal rank 1. Can we decide inpolynomial time whether P contains an integer point?
It does not seem straightforward to use the Chvatal rank 1 condition, because it is NP-hard todecide whether a rational polytope has Chvatal rank 1 (Cornuejols and Li [9]). We also note thatthe Chvatal rank of a polyhedron is not directly related to its geometry. In particular, the Chvatalrank is not invariant under translation. The following example shows that the Chvatal rank of apolyhedron may vary significantly under translation.
Example 2. Let Q1 := {x ∈ [0, 1]n :∑n
j=1 vj(1− xj) + (1− vj)xj ≥12 ∀v ∈ {0, 1}n}. Notice that
Q1 contains no integer point. Chvatal, Cook and Hartmann [7] proved that the Chvatal rank ofQ1 is exactly n. Now, let us translate Q1 so that its center point is at the origin. We denote by Q2
the resulting polytope. Since Q2 ⊆ [−12 ,
12 ]
n, the only integer point contained in Q2 is the origin.We can obtain both xi ≥ 0 and xi ≤ 0 as Chvatal inequalities for Q2 for all i = 1, · · · , n. Hence,the Chvatal rank of Q2 is exactly 1.
The difficulty in understanding the Chvatal rank 1 condition is an indication that the aboveopen question might not be easy to answer in general. Next, we present several special cases ofOpen Question 1 where the answer is positive.
2.1 Satisfiability problem with Chvatal rank 1
The satisfiability problem is NP-complete (see [17]), and it can be transformed to a binary integerprogram. Given a formula in conjunctive normal form with m clauses that consist of literalsx1, · · · , xn and their negations, we can turn this formula into a polytope in the unit hypercube.For example, the formula (x1 ∨¬x2 ∨x3)∧ (¬x3 ∨x4) can be turned into the polytope {x ∈ [0, 1]4 :x1 + (1 − x2) + x3 ≥ 1, (1 − x3) + x4 ≥ 1}. The linear description of such a polytope consists ofgeneralized set covering inequalities and the bounds 0 ≤ x ≤ 1. We call it a SAT polytope. Noticethat the satisfiability problem is equivalent to the integer feasibility problem over a SAT polytope.
Open Question 2. Given a SAT polytope P whose Chvatal rank is 1, can we decide in polynomialtime whether P contains an integer point?
The k-satisfiability problem is a variant of the satisfiability problem where each clause in agiven formula has at most k literals. It remains NP-complete for k ≥ 3 (see [17]). There is a simplepolynomial algorithm for the 2-satisfiability problem. We consider Open Question 2 for the casewhere each clause contains at least 3 literals.
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Theorem 3. Let P be a SAT polytope such that each generalized set covering inequality has atleast 3 variables. If P has Chvatal rank 1, then P always contains an integer point.
Proof. Observe that setting any variable to 0 or 1, and all other n − 1 variables to 1/2 satisfiesall the constraints of P (because every generalized set covering inequality involves at least threevariables). In other words, the middle point of each facet of the hypercube [0, 1]n is contained in P .A theorem of Chvatal, Cook and Hartmann [7] implies that the Chvatal closure of P contains themiddle point (12 , · · · ,
12 ) of the hypercube, so the Chvatal closure of P is always nonempty. Because
the Chvatal rank of P is 1, P contains an integer point.
The gap between Open Question 2 and Theorem 3 is on the SAT polytopes defined by theboolean formulas involving both clauses with 2 literals and clauses with at least 3 literals. SATpolytopes whose generalized set covering inequalities have at most 2 variables are well understood byGerards and Schrijver [18]. They gave a characterization of the Chvatal closure of a SAT polytopein such a case, and they provided a polynomial algorithm to separate over it. Furthermore, weremark that the Chvatal rank of a SAT polytope in that case is always 1 whenever it contains nointeger point. However, the Chvatal closure of a SAT polytope which includes both generalized setcovering inequalities with 2 variables and 3 variables has not been studied.
2.2 When a few Chvatal cuts are sufficient
In this section, we consider another special case of Open Question 1, where we assume that theinteger hull of a given polyhedron can be obtained by adding a constant number of Chvatal in-equalities.
Open Question 3. Let P ⊂ Rn be a rational polyhedron such that PI can be obtained by addingat most k Chvatal inequalities to the description of P , for some constant k. Can we solve theinteger feasibility problem of P in polynomial time?
In fact, Open Question 3 is open even when k = 1. The difficulty might be due to the resultthat it is NP-complete to decide whether PI can be obtained from a polytope P by adding oneChvatal inequality, which will be proved in Section 6.
Figure 1: When one Chvatal inequality is sufficient in R2
Remark 4. Let P = {x ∈ Rn : Ax ≤ b} be a rational polytope such that addition of one Chvatalinequality to P gives PI . Then there exists an algorithm for the integer feasibility problem over P
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which runs in time bounded by mnn3poly(L) where m and L denote the number of constraints in Pand the encoding size of P , respectively.
This is easy to show because if v is a fractional vertex of P , then v should be removed by the Chvatalinequality. Therefore P contains an integer point if and only if an extreme point of P is integral.In this case, there is a trivial algorithm for the integer feasibility problem: check every vertex ofP and conclude that PI 6= ∅ if there exists an integral vertex or PI = ∅ otherwise. Since thereare O(mn) extreme points of P and the time complexity of the Gaussian elimination method isbounded by n3poly(L), the algorithm runs in time bounded by mnn3poly(L). In fact, a byproductof Theorem 18 shows the existence of a 2O(n)poly(L) time algorithm for the case of k = 1.
In the following, we are able to prove that Open Question 3 is true in a special case.
Theorem 5. Let k be a positive constant in Z. Given a rational simplex P ⊂ Rn such that PI
can be obtained from P by adding at most k Chvatal inequalities, and a vector w ∈ Qn, there is apolynomial-time algorithm to optimize the linear function wx over PI .
Proof. Suppose that the dimension of P is ℓ for some ℓ ≤ n. Let P be written as {x ∈ Rn : Ax =b, Cx ≤ d}, where Ax = b is the system of all implicit equalities. We denote by Ex ≤ f the set ofk Chvatal inequalities such that PI = {x ∈ Rn : Ax = b, Cx ≤ d, Ex ≤ f}. So the inequalitiesEx ≤ f + ǫ1 are valid for P , where ǫ ∈ (0, 1) and 1 denotes the all-1 vector, and P ⊆ S, whereS := {x ∈ Rn : Ax = b, Ex ≤ f + ǫ1}.
We can find in polynomial time a unimodular matrix U such that AU = (D, 0) is a Hermitenormal form of A. If D−1b is not integral, we can just conclude that P does not contain an integerpoint. Thus, we may assume that D−1b is integral. Let U1 and U2 denote the two submatricesof U which consist of the first n − ℓ columns of U and the last ℓ columns of U , respectively. Letu : Rn → Rn be a unimodular transformation defined by u(x) = U−1x. Consider the images of P ,PI , and S under u:
u(P ) ={
(y1, y2) ∈ R(n−ℓ)+ℓ : y1 = D−1b, CU2y2 ≤ d− CU1D−1b
}
,
u(PI) ={
(y1, y2) ∈ R(n−ℓ)+ℓ : y1 = D−1b, CU2y2 ≤ d− CU1D−1b, EU2y2 ≤ f − EU1D
−1b}
,
u(S) ={
(y1, y2) ∈ R(n−ℓ)+ℓ : y1 = D−1b, EU2y2 ≤ f + ǫ1− EU1D−1b
}
.
Note that u(P ) is an ℓ-dimensional simplex in Rn, so Q :={
y2 ∈ Rℓ : CU2y2 ≤ d− CU1D−1b
}
is an ℓ-dimensional simplex in Rℓ. Furthermore, u(PI) is integral. Since D−1b is integral, {y2 ∈Rℓ : CU2y2 ≤ d − CU1D
−1b, EU2y2 ≤ f − EU1D−1b} is integral and thus Q ∩ {y2 ∈ Rℓ :
EU2y2 ≤ f − EU1D−1b} is integral. We claim that the inequalities in the system EU2y2 ≤
f − EU1D−1b are Chvatal inequalities for Q. In fact, we know that u(P ) ⊆ u(S), so Q ⊆
{
y2 ∈ Rℓ : EU2y2 ≤ f + ǫ1− EU1D−1b
}
. That means the inequalities in EU2y2 ≤ f + ǫ1 −EU1D
−1b are all valid for Q, so those in the system EU2y2 ≤ f − EU1D−1b are Chvatal in-
equalities for Q. Now, we have obtained a full-dimensional rational simplex Q in Rℓ such that itsinteger hull QI can be described by adding at most k Chvatal inequalities.
Q has ℓ+1 constraints in its description, so QI can be described by ℓ+k+1 linear inequalities.When ℓ ≤ k, the dimension of Q is fixed and we can optimize a linear function over QI in polynomialtime by Lenstra’s algorithm [25]. Thus, we may assume that ℓ > k. Suppose that QI is not empty.Then let z ∈ Zℓ be an extreme point of QI . So there are ℓ linearly independent inequalities in thedescription of QI that are active at z. This means that at least ℓ− k inequalities among the ℓ+ 1inequalities in the original description of Q are active at z. Thus, z belongs to a k-dimensional face
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of Q. Hence, if no k-dimensional face of Q contains an integer point, QI is empty. Since k is fixed,we can optimize a linear function over the integer hull of each k-dimensional face of Q. Notice thatthere are exactly
(ℓ+1k+1
)
k-dimensional faces of Q. Therefore, we can optimize a linear function overQI in polynomial time.
Theorem 5 suggests the following open question.
Open Question 4. Given a rational simplex P ⊂ Rn with Chvatal rank 1, can we decide inpolynomial time whether P contains an integer point?
2.3 Rounded polytopes
A full-dimensional polytope P ⊂ Rn is rounded with factor ℓ > 1 if Bn2 (a, r) ⊆ P ⊆ Bn
2 (a, ℓr),where Bn
2 (p, q) denotes an Euclidean ball {x ∈ Rn : ‖x − p‖2 ≤ q} which is centered at p withradius q. In this section, we prove the following theorem.
Theorem 6. Let ℓ > 1 be a constant, and let P ⊂ Rn be a rounded polytope with factor ℓ. If PI
can be obtained by adding one Chvatal inequality to the description of P , then we can decide inpolynomial time whether P contains an integer point.
To prove this theorem, we use the notion of integer width of a convex set.
Definition 7. Let K ⊂ Rn be a convex set and d ∈ Zn. The integer width of K along d is
w(K, d) := ⌊sup{dx : x ∈ K}⌋ − ⌈inf{dx : x ∈ K}⌉+ 1.
The integer width of K, w(K,Zn), is the infimum of the values w(K, d) over all d ∈ Zn \ {0}.
w(K,Zn) := infd∈Zn\{0}
w(K, d).
Lemma 8. Let P ⊂ Rn be a rounded polytope with factor ℓ > 1. If there exists an integral vector csuch that w(P, c) ≤ k for some nonnegative integer k, then either ‖c‖2 ≤ (k + 1)ℓ or w(P, ei) ≤ 1for all i = 1, · · · , n.
Proof. Since P is rounded with factor ℓ, P satisfies Bn2 (a, r) ⊆ P ⊆ Bn
2 (a, ℓr) for some r > 0and a ∈ Rn. Assume that ‖c‖2 > (k + 1)ℓ. Since w(P, c) ≤ k, there exists c0 ∈ Z such thatc0 < cx < c0 + k + 1 for all x ∈ P . Since Bn
2 (a, r) ⊆ P ⊆ {x ∈ Rn : c0 < cx < c0 + k + 1} andthe distance between the two hyperplanes {x ∈ Rn : cx = c0} and {x ∈ Rn : cx = c0 + k + 1}is exactly (k + 1)/‖c‖2, 2r is at most (k + 1)/‖c‖2. Hence, we get r ≤ k+1
2‖c‖2< 1
2ℓ , i.e., 2ℓr < 1.
Suppose that there is some i such that w(P, ei) ≥ 2. Then there are two points u, v ∈ P such thatui ≤ b and vi ≥ b + 1 for some b ∈ Z. So ‖u − v‖2 ≥ |ui − vi| ≥ 1. Since Bn
2 (a, ℓr) contains P ,the distance between any two points in P is at most 2ℓr and thus we get 2ℓr ≥ 1. However, thiscontradicts the above result that 2ℓr < 1. Therefore, w(P, ei) ≤ 1 for all i = 1, · · · , n.
Proof of Theorem 6. Consider the following algorithm:
(1) For each c ∈ Zn with ‖c‖2 ≤ ℓ, compute w(P, c). If w(P, c) = 0 for some c with ‖c‖2 ≤ ℓ,then PI = ∅. Otherwise, go to step (2).
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(2) Compute w(P, ei) for i ∈ [n]. If there exists i ∈ [n] such that w(P, ei) ≥ 2, then PI 6= ∅. Ifthere exists i ∈ [n] such that w(P, ei) = 0, then PI = ∅. Otherwise, go to step (3).
(3) Let zj := ⌊max{xj : x ∈ P}⌋ for j = 1, · · · , n. If (z1, · · · , zn) ∈ P , then PI 6= ∅. Otherwise,PI = ∅.
Step (1) can be done in polynomial time, because there are at most(nℓ
)
2ℓ(2ℓ−1
ℓ
)
integral vectors cwith ‖c‖2 ≤ ℓ.
By assumption, there exists a Chvatal inequality dx ≤ d0 such that {x ∈ P : dx ≤ d0} = PI .Note that PI is empty if and only if w(P, d) = 0. Going into Step (2), we have w(P, c) ≥ 1 for allc ∈ Zn with ‖c‖2 ≤ ℓ. If w(P, ei) ≥ 2 for some i ∈ [n], then w(P, d) ≥ 1 by Lemma 8 (when k = 0)and thus PI 6= ∅. If w(P, ei) = 0 for some i ∈ [n], then PI is trivially empty.
Therefore, going into Step (3), we have w(P, ei) = 1 for all i = 1, · · · , n, and P can have atmost one integer point. This point z can be computed by solving n linear programs, therefore, inpolynomial time.
Remark 9. The integer feasibility problem can be solved in quasi-polynomial time, when P ⊆ Rn isa rounded polytope with factor O(log n) and its integer hull can be obtained by adding one Chvatalinequality. In fact, the algorithm given in the proof of Theorem 6 solves nO(logn) linear programsin this case.
3 Flatness theorem for closed convex sets with empty Chvatal
closure
Recall the definition of integer width of a convex set K given in Definition 7. When K is unboundedor has a large volume, there exists a direction d ∈ Zn \ {0} where w(K, d) is large. On the otherhand, it is possible that there is a direction d ∈ Zn \ {0} such that w(K, d) is relatively small if Kdoes not contain any integer point. In fact, the famous flatness theorem by Khinchine [24] statesthat w(K,Zn) for any compact convex set K containing no integer point is bounded by f(n), whichis only a function of the dimension n. Khinchine’s flatness theorem [24] shows that f(n) = (n+1)!.A crucial component of Lenstra’s algorithm [25] is to find a flat direction d ∈ Zn\{0} of a polyhedronP ⊂ Rn containing no integer point. Lenstra [25] gave a polynomial algorithm to find a directiond ∈ Zn \ {0} such that w(K, d) ≤ 2O(n2) for a given lattice-free compact convex set K. Then, itgenerates 2O(n2) subproblems in Rn−1 by intersecting K with 2O(n2) parallel hyperplanes orthogonalto d. Hence, the algorithm works recursively, and the number of total steps required is 2O(n3).
Over the last few decades there have been huge improvements on the upper bound f(n) (see [1,2, 23, 24, 29]). The current best known asymptotic upper bound is f(n) = O(n4/3polylog(n)) givenby Banaszczyk, Litvak, Pajor, and Szarek [2] and Rudelson [29]. It has been even conjectured thatf(n) = O(n). However, the existence of a polynomial algorithm to find a direction d ∈ Zn suchthat w(K, d) = O(n4/3polylog(n)) for a convex set K containing no integer point is not known.Dadush, Peikert and Vempala [12] and Dadush and Vempala [13] developed an algorithm to findall vectors d ∈ Zn \ {0} such that w(K, d) = w(K,Zn) in 2O(n)poly(L) time and space.
In this section, we prove that f(n) is indeed n if K is a closed convex set with empty Chvatalclosure. The Chvatal closure of a closed convex set is defined similarly to that of a polyhedron.
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Definition 10. Let K ⊂ Rn be a closed convex set. The Chvatal closure of K is defined as
K ′ :=⋂
c∈Zn
{x ∈ Rn : cx ≤ ⌊sup{cx : x ∈ K}⌋} .
If K is a compact convex set, its Chvatal closure is a rational polytope [11]. The following is themain result of this section.
3.1 Flatness result
Theorem 11. The integer width of any closed convex set in Rn whose Chvatal closure is empty isat most n.
The upper bound given by Theorem 11 turns out to be quite tight as shown in the followingproposition.
Proposition 12. There exists a polytope in Rn such that its Chvatal closure is empty and itsinteger width is n− 1.
Proof. Let Pn : {x ∈ Rn : x ≥ 1n+11,
∑ni=1 xi ≤ n − 1 + n
n+1}. Then Pn is the convex hull of
(n− 1)ei + 1n+11 for i = 1, · · · , n and 1
n+11. Since xi ≥ 1 is valid for P ′n for each i,
∑ni=1 xi ≥ n is
valid for P ′n. Together with
∑ni=1 xi ≤ n− 1 + n
n+1 , this shows the emptiness of P ′n.
Figure 2: P2 in R2
Now we show that the integer width of Pn is n − 1. Let d ∈ Zn \ {0}. Since the inte-ger width of Pn along d is the same as that along −d, we may assume
∑ni=1 di ≥ 0. No-
tice that max{dx : x ∈ Pn} = (n − 1)max{d1, · · · , dn} + 1n+1
∑ni=1 di and min{dx : x ∈
Pn} = (n − 1)min{0, d1, · · · , dn} + 1n+1
∑ni=1 di. Then the integer width of Pn along d is either
(n−1)(max{d1, · · · , dn}−min{0, d1, · · · , dn}) or (n−1)(max{d1, · · · , dn}−min{0, d1, · · · , dn})+1.Clearly, max{d1, · · · , dn} −min{0, d1, · · · , dn} is at least 1. Hence, the integer width of Pn along dis at least n− 1. It is easy to show that the integer width of Pn along 1 is exactly n− 1.
As a corollary of Theorem 11, we obtain an improved flatness theorem for polyhedra withChvatal rank 1. The result will be used in developing an algorithm for solving the integer feasibilityproblem over the rational polyhedra with Chvatal rank 1 in the later part of this section.
Corollary 13. Let P ⊂ Rn be a polyhedron with Chvatal rank 1. Then, either P contains aninteger point or the integer width of P is at most n.
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3.2 Proof of Theorem 11
To prove Theorem 11, we introduce the concept of a simplicial cylinder. Let P ⊂ Rn be a full-dimensional rational polyhedron. We denote by L and L⊥ the lineality space of P and its comple-ment, respectively. We say that P is a simplicial cylinder if P ∩ L⊥ is a simplex.
Remark 14. A simplicial cylinder P ⊂ Rn whose lineality space L has dimension n − ℓ can bedescribed by ℓ+ 1 linear inequalities.
Given a full-dimensional rational polyhedron P ⊂ Rn and its linear description Ax ≤ b, whereeach entry in [A; b] is an integer and each row of A has relatively prime integers, we call P a thinsimplicial cylinder if it is a simplicial cylinder and Ax ≤ b− 1, where 1 denotes the all 1 vector, isan infeasible system.
Figure 3: Thin simplicial cylinders in R2
Remark 15. The Chvatal closure of any compact convex set contained in the interior of a thinsimplicial cylinder is empty.
Theorem 16. Any closed convex set whose Chvatal closure is empty is contained in the interiorof a thin simplicial cylinder.
Proof. Let K ⊂ Rn be a closed convex set whose Chvatal closure is empty. Helly’s theorem impliesthat there be ℓ + 1 Chvatal inequalities of K for some ℓ ≤ n such that the intersection of thecorresponding linear half-spaces is empty. In other words, there exists a system Ax ≤ b − ǫ1 ofℓ+ 1 linear inequalities valid for K
a1x ≤ b1 − ǫa2x ≤ b2 − ǫ
...aℓx ≤ bℓ − ǫ
aℓ+1x ≤ bℓ+1 − ǫ
where 0 < ǫ < 1, a1, · · · , aℓ+1 ∈ Zn, and b1, · · · , bℓ+1 ∈ Z for some ℓ ≤ n such that the systemAx ≤ b−1 obtained by rounding down the right hand side values of the above is infeasible. Withoutloss of generality, we may assume that each ai has relatively prime entries. We may also assume
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that the system is minimal in a sense that {x ∈ Rn : aix ≤ bi − 1 for i ∈ I} is not empty for anyproper subset I of [ℓ+ 1].
Consider the polyhedron P := {x ∈ Rn : Ax ≤ b}. We claim that its recession cone C := {x :Ax ≤ 0} has empty interior. Otherwise, the polyhedron P contains points in the form of x+ kr forsome x ∈ P and some ray vector r ∈ Rn in the interior of C, where k ∈ R+. For k large enough,the points of the form are also in the polyhedron Q := {x ∈ Rn : Ax ≤ b− 1}, which is empty, acontradiction. Therefore, the linear space C − C has dimension strictly less than n.
By the Minkowski-Weyl theorem, we can write the polyhedron P as P = S + C where Sis a polytope. Consider the cylinder R := S + C − C. Consider all the inequalities aix ≤ bi,i = 1, · · · , t, in the description of P that are valid for R. Then every ray vector r ∈ C satisfiesair < 0 for i = t + 1, · · · , ℓ + 1. We claim that the linear system aix ≤ bi − 1, i = 1, · · · , t, isinfeasible. If aix ≤ bi − 1, i = 1, · · · , t, were feasible, then, by the same argument as given inthe above paragraph, Q would be nonempty, a contradiction. Thus aix ≤ bi − 1, i = 1, · · · , t, isinfeasible. By the minimality of the system, this implies t = ℓ+ 1, and therefore S is a simplex ofdimension ℓ.
Proof of Theorem 11. Let K ⊂ Rn be a closed convex set whose Chvatal closure is empty. ByTheorem 16, there exists a thin simplicial cylinder P := {x ∈ Rn : Ax ≤ b} containing K in itsinterior. Hence, there exists a small ǫ ∈ (0, 1) such that K ⊆ {x ∈ Rn : Ax ≤ b− ǫ1}. We assumethat the number of rows in A is ℓ + 1 for some ℓ ≤ n. We denote by a1, · · · , aℓ+1 the rows of A.Notice that P ∩ L⊥ is an ℓ-dimensional simplex, where L and L⊥ denote the lineality space of Pand its orthogonal complement, respectively.
We will show that the integer width of P along some ai is at most ℓ+1. Then the integer widthof K is at most ℓ, because the hyperplane defined by aix = bi does not go through K. Supposethat the integer width of P along each ai is at least ℓ + 2 for the sake of contradiction. Then,the width of P along each ai is at least ℓ + 1. Using an affine transformation, we can transformP to {x ∈ Rn : x1, · · · , xℓ ≥ 0,
∑ℓi=1 xi ≤ 1}. Under the same affine transformation, we know
that {x ∈ Rn : Ax ≤ b − 1} is transformed to {x ∈ Rn : xi ≥ ǫi ∀ i ∈ [ℓ],∑ℓ
i=1 xi ≤ 1 − ǫ}for some ǫi ≤ 1
l+1 for i ∈ [ℓ] and ǫ ≤ 1l+1 . Notice that 1
l+11 is contained in {x ∈ Rn : xi ≥
ǫi ∀ i ∈ [ℓ],∑ℓ
i=1 xi ≤ 1 − ǫ}. However, {x ∈ Rn : Ax ≤ b− 1} is empty by the assumption thatP is a thin simplicial cylinder, and it cannot be transformed to a nonempty set under any affinetransformation. With this contradiction, we have proved that the integer width of K is at mostℓ ≤ n.
In the following, we give a simpler proof of the same flatness theorem for bounded closed convexsets.
Proposition 17. Let K ⊂ Rn be a compact convex set whose Chvatal closure is empty. If E(C, a) ⊂K ⊂ E(1ℓC, a), then the integer width of K is at most ⌈ℓ⌉.
Proof. Since the Chvatal closure of K is empty, the center point a of the inner ellipsoid shouldbe cut off by a Chvatal inequality of K. In other words, there exists (d, d0) ∈ Zn+1 such thatmax{dx : x ∈ K} < d0 and da > d0 − 1. Let x∗ := argmax{dx : x ∈ E(C, a)}. Thendx∗ < d0. Geometrically, the minimum of dx over E(1ℓC, a) should be obtained at a − ℓ(x∗ − a).Then min{dx : x ∈ K} ≥ min{dx : x ∈ E(1ℓC, a)} = d(a− ℓ(x∗ − a)) > d0 − ℓ− 1. Therefore, theinteger width of K along d is at most ⌈ℓ⌉.
10
The Lowner-John ellipsoid can be used to show that there always exists such a pair of ellipsoidswith ℓ ≤ n. Hence, the proposition also implies that the integer width of every compact convex setin Rn whose Chvatal closure is empty is at most n.
3.3 A Lenstra-type algorithm
Recently Hildebrand and Koppe [21], Dadush, Peikert, and Vempala (see [10, 12, 13]) improvedLenstra-type algorithms for integer programming. Their algorithms are similar to Lenstra’s al-gorithm in spirit in that a main step consists in finding a flat direction of a lattice-free convexbody. In particular, Dadush, Peikert, and Vempala (see [10, 12, 13]) used a 2O(n)poly(L) timealgorithm to find a flattest direction for a convex body containing no integer point. Together withRudelson’s flatness theorem [29], they proved that the time complexity of the algorithm is boundedby 2O(n)
(
n4/3polylog(n))n
poly(L). Theorem 11 implies that there exists a 2O(n)nnpoly(L) timeLenstra-type algorithm for the integer feasibility problem over Chvatal rank 1 rational polyhedra.On the other hand, Proposition 12 indicates that we cannot improve this time complexity if we usea Lenstra-type procedure. Note that this does not improve the current best algorithm for integerprogramming. Dadush [10] provided a 2O(n)nnpoly(L) time Kannan-type algorithm for integerprogramming over general convex compact sets in his doctoral dissertation, and we remark that itis the fastest algorithm for integer programming. Instead of finding one flat direction at a time, hisalgorithm finds many flat directions at each step, thereby reducing the number of recursive stepsfrom
(
n4/3polylog(n))n
to (3n)n.Based on Theorem 11 and Theorem 16, we can state the following theorem. We do not describe
our algorithm in this paper, because it is similar to the earlier work done by Dadush, Peikert, andVempala (see [10, 12, 13]).
Theorem 18. Let P ⊂ Rn be rational polyhedron with Chvatal rank 1. Assume that if P containsno integer point, then P is contained in the interior of a thin simplicial cylinder defined by ℓ + 1inequalities. Then, there exists a 2O(n)ℓnpoly(L) time Lenstra-type algorithm that decides whetherP contains an integer point, where L is the encoding size of P .
Since any rational polyhedron with empty Chvatal closure in Rn is always contained in theinterior of a thin simplicial cylinder which is defined by at most n + 1 inequalities, Theorem 18directly implies the following.
Corollary 19. There is a Lenstra-type algorithm that can decide in 2O(n)nnpoly(L) time, where Lis the encoding size of P , whether a given rational Chvatal rank 1 polyhedron P ⊂ Rn contains aninteger point.
4 Deciding emptiness of the Chvatal closure
In this section, we prove that deciding whether a given polytope has empty Chvatal closure isNP-complete, even if the input polytope is either a polytope contained in the unit hypercube or asimplex. Consequently, the optimization and separation problems over the Chvatal closure remainNP-hard, when the input is either a rational polytope contained in the unit hypercube or a rationalsimplex. This extends the result of Eisenbrand [16] that the optimization and separation problemsover the Chvatal closure of a polyhedron are NP-hard.
11
Equality Knapsack Problem (see [17]). Given positive integers a1, · · · , an, b, is there a set ofnonnegative integers {xi}
ni=1 satisfying
∑ni=1 aixi = b.
Without loss of generality, we assume that a1, · · · , an are relatively prime and that a1 < · · · < an.Throughout this section, we refer to the equality knapsack problem as “the knapsack problem” forsimplicity. Using the input data for an instance of the knapsack problem, we construct a rationalpolytope in the unit hypercube and a rational simplex. By doing so, we are able to provide apolynomial-time reduction from the knapsack problem to the problem of deciding whether theChvatal closure is empty.
4.1 The case of polytopes contained in the unit hypercube
In this section, we prove NP-completeness of the following problem.
Problem 20. Given a rational polytope P contained in the n-dimensional unit hypercube [0, 1]n,is the Chvatal closure of P empty?
First, we will construct a polynomial reduction from the knapsack problem to Problem 20.Let a1, · · · , an−4, b be positive integers that satisfy 1 ≤ a1, · · · , an−4 < b. Let a rational polytopeP1 ⊆ [0, 1]n be defined as the convex hull of the following n+ 6 vectors v1, · · · , vn+6 ∈ [0, 1]n:
v1 := ( 12b , 0, · · · , 0, 0, 0, 1
2b , 0, 0 )v2 := ( 0, 1
2b , · · · , 0, 0, 0, 12b , 0, 0 )
...vn−4 := ( 0, 0, · · · , 0, 1
2b , 0, 12b , 0, 0 )
vn−3 := ( 0, 0, · · · , 0, 0, 0, 1/2, 1/2, 1/2 )vn−2 := ( 1, 1, · · · , 1, 1, 1, 1/2, 1/2, 1/2 )vn−1 := ( 1/2, 1/2, · · · , 1/2, 1/2, 1/2, 1, 1, 1 )vn := ( 1/4, 1/4, · · · , 1/4, 1/4, 1/4, 1/4, 1/4, 1/4 )
vn+1 := ( 1/2, 1/2, · · · , 1/2, 1/2, 1/2, 1, 1, 1/2 )vn+2 := ( 1/2, 1/2, · · · , 1/2, 1/2, 1/2, 0, 0, 1/2 )vn+3 := ( 1/2, 1/2, · · · , 1/2, 1/2, 1/2, 1/2, 1, 1 )vn+4 := ( 1/2, 1/2, · · · , 1/2, 1/2, 1/2, 1/2, 0, 0 )vn+5 := ( a1
2b ,a22b , · · · , an−5
2b , an−4
2b , 0, 0, 12 −
14b , 0 )
vn+6 := ( 1− a12b , 1− a2
2b , · · · , 1− an−5
2b , 1− an−4
2b , 1, 12 +
14b , 0, 0 )
Lemma 21. A linear description of P1 can be obtained in polynomial time, and P1 contains nointeger point.
Proof. We can easily show that P1 is a full-dimensional polytope, so the number of facets of P1 isat most
(n+6n
)
≤ n6. Given n affinely independent points among v1, · · · , vn+6, we can compute thehyperplane containing theses n points using the Gaussian elimination method. Since the encodingsize of each vi is polynomially bounded by log a1, · · · , log an−4, log b, and n, the complexity of thehyperplane is also polynomially bounded by the input encoding size. Therefore, we can polynomiallyfind each facet of P1. Lastly, P1 := conv{v1, · · · , vn+6} ⊂ [0, 1]n contains no integer point, becauseevery point in {v1, · · · , vn+6} is fractional.
12
Let Q1 := {x ∈ Zn−4+ :
∑n−4i=1 aixi = b}. The following two lemmas show that P ′
1 = ∅ if andonly if Q1 6= ∅, thereby providing a polynomial reduction from the knapsack problem to Problem20.
Lemma 22. If Q1 6= ∅, then P ′1 = ∅.
Proof. Let (w1, · · · , wn−4) ∈ Q1. Then∑n−4
i=1 aiwi = b and wi ≥ 0 for i = 1, · · · , n − 4. Letd := (w1, · · · , wn−4,−
∑n−4i=1 wi, 1,−1, 1) ∈ Zn. Notice that wk ≤ akwk ≤
∑n−4i=1 aiwi = b, so we get
wk
2b ≤ 12 . Since b > 1, we know that 0 < 1
2b ≤ 14 . Thus, 0 < dvk = wk
2b + 12b < 1 for k = 1, · · · , n− 4.
It is easy to show that dvn−3 = dvn−2 = dvn+1 = dvn+2 = dvn+3 = dvn+4 = 12 , dv
n = 14 , and
dvn−1 = 1. In addition, dvn+5 = dvn+6 = 14b . That means 0 < dvi < 1 for i 6= n−1 and dvn−1 = 1.
We know that dx > 0 is valid for P1, because dvk > 0 for k = 1, · · · , n+6 and P1 is the convexhull of v1, · · · , vn+6. This implies dx ≥ 1 is valid for P ′
1. In fact, P1 ∩{x ∈ Rn : dx ≥ 1} = {vn−1}.Notice that xn−3 + xn−2 + xn−1 + xn ≤ 7
2 is also valid for P1, because it is valid for every vk fork = 1, · · · , n+6. Then xn−3+xn−2+xn−1+xn ≤ 3 is valid for P ′
1, and vn−1 violates this inequality.Therefore, P1 ∩ {x ∈ Rn : dx ≥ 1, xn−3 + xn−2 + xn−1 + xn ≤ 3} is empty. Hence, P ′
1 = ∅.
Lemma 23. If P ′1 = ∅, then Q1 6= ∅.
Proof. Let u := 12v
n−3 + 12v
n−2. Then u = (12 ,12 , · · · ,
12 ,
12 ,
12 ,
12 ,
12 ,
12). Since P ′
1 = ∅, there is aChvatal inequality that is violated by u. In other words, there exists a valid inequality dx ≤ d0+αfor P1 such that (d, d0) ∈ Zn+1, 0 < α < 1, and du > d0. We claim that d and d0 satisfy thefollowing five properties:
1) dn−3 = −∑n−4
i=1 di.
2) d0 = −1.
3) dn−2 = dn = −1 and dn−1 = 1.
4)∑n−4
i=1 aidi = −b.
5) di ≤ 0 for i = 1, · · · , n− 4.
Then, (−d1, · · · ,−dn−4) ∈ Q1, so Q1 6= ∅.Since d0 < du ≤ d0 + α < d0 + 1, we get d0 < 1
2
∑ni=1 di < d0 + 1. In addition, we know that
dvk ≤ d0+α < d0+1 for k = 1, · · · , n+6. The integrality of∑n
i=1 di implies that 12
∑ni=1 di should
be equal to d0+12 , and thus we get
∑ni=1 di = 2d0+1 and du = d0+
12 . Consider dv
n−3 and dvn−4:
d0 + 1 > dvn−3 = du−1
2
n−3∑
i=1
di = d0 +1
2−
1
2
n−3∑
i=1
di, (1)
d0 + 1 > dvn−2 = du+1
2
n−3∑
i=1
di = d0 +1
2+
1
2
n−3∑
i=1
di. (2)
By (1) and (2), we get −1 <∑n−3
i=1 di < 1. Since∑n−3
i=1 di is an integer,∑n−3
i=1 di = 0 and the firstproperty is satisfied. Then we know that dn−2 + dn−1 + dn = 2d0 + 1. Now, consider dvn−1 anddvn:
d0 + 1 > dvn−1 = du+1
2(dn−2 + dn−1 + dn) = 2d0 + 1, (3)
13
d0 + 1 > dvn =1
2du =
1
2d0 +
1
4. (4)
By (3) and (4), we obtain −32 < d0 < 0 and thus d0 = −1. So the second property holds and
dn−2 + dn−1 + dn = −1. Consider dvn+1 and dvn+2:
d0 + 1 > dvn+1 = du+1
2(dn−2 + dn−1) = d0 +
1
2+
1
2(dn−2 + dn−1), (5)
d0 + 1 > dvn+2 = du−1
2(dn−2 + dn−1) = d0 +
1
2−
1
2(dn−2 + dn−1). (6)
By (5) and (6), we know that −1 < dn−2 + dn−1 < 1. So, dn−2 + dn−1 = 0. Similarly, we getdn−1+dn = 0 by considering dvn+3 and dvn+4. Together with the observation dn−2+dn−1+dn = −1,we get dn−1 = 1 and dn−2 = dn = −1. Hence, the third property is satisfied. To prove the fourthproperty, we consider dvn+5 and dvn+6:
dvn+5 =1
2b
n−4∑
i=1
aidi + (1
2−
1
4b) < d0 + 1 = 0, (7)
which implies that∑n−4
i=1 aidi < −b+ 12 , so
∑n−4i=1 aidi ≤ −b since the sum is an integer;
dvn+6 =
n−3∑
i=1
di −1
2b
n−4∑
i=1
aidi − (1
2+
1
4b) = −
1
2b
n−4∑
i=1
aidi − (1
2+
1
4b) < d0 + 1 = 0, (8)
which implies that∑n−4
i=1 aidi > −b− 12 , so
∑n−4i=1 aidi ≥ −b since the sum is an integer. Therefore,
∑n−4i=1 aidi = −b. Lastly, consider dvk for k = 1, · · · , n− 4:
dvk =1
2bdk −
1
2b< d0 + 1 = 0. (9)
By (9), dk < 1 and thus dk ≤ 0.
We are now ready to state the first main result of this section.
Theorem 24. Deciding emptiness of the Chvatal closure of a rational polytope contained in theunit hypercube is NP-complete.
Proof. First, deciding emptiness of the Chvatal closure of a rational polyhedron is in NP [9]. ByLemma 21, Lemma 22 and Lemma 23, we also know that there is a polynomial reduction from theknapsack problem to Problem 20. Thus, the decision problem is NP-complete.
4.2 The case of simplices
A full-dimensional polytope P ⊂ Rn is called a simplex if it is the convex hull of n + 1 affinelyindependent points. The integer programming feasibility problem over a rational simplex is NP-complete because of the following polynomial reduction of the knapsack problem to it [32]: considerpositive integers a1, · · · , an, b. Let vi := b
aiei where ei denotes the ith unit vector for i = 1, · · · , n.
Let vn+1 :=b− 1
2
n ( 1a1, · · · , 1
an). Let conv{v1, · · · , vn+1} denote the convex hull of v1, · · · , vn+1.
Note that avn+1 = b − 12 and avi = b for i = 1, · · · , n. Then, conv{v1, · · · , vn+1} ∩ Zn =
conv{v1, · · · , vn} = {x ∈ Zn :∑n
i=1 aixi = b, x ≥ 0}.In this section, we prove NP-completeness of the following problem.
14
Problem 25. Given a rational simplex P ⊂ Rn, is the Chvatal closure of P empty?
Let a1, · · · , an−1, b be positive integers satisfying 1 ≤ a1, · · · , an−1 < b. Using the givena1, · · · , an−1, b, we define the following n+ 1 vectors in Rn:
v1 := ( 12rB , 0, · · · , 0, 1
2r −b
2rBA )
v2 := ( 0, 12rB , · · · , 0, 1
2r −b
2rBA )...
vn−1 := ( 0, 0, · · · , 12rB , 1
2r −b
2rBA )vn := ( ra1, ra2, · · · , ran−1 −rb+ 1
2 )vn+1 := ( −ra1, −ra2, · · · , −ran−1 rb+ 1 )
where A and B denote∑n−1
i=1 ai and the smallest integer greater than bA , respectively and r :=
2016b + 12b . Let P2 := conv{v1, · · · , vn+1} ⊂ Rn. It is straightforward to show that P2 is an
n-dimensional simplex.
Lemma 26. One can polynomially find a linear description of P2.
Proof. Since P2 is a full-dimensional rational simplex in Rn, it contains exactly n+1 facets. One canobtain each facet-defining inequality of P2 by the Gaussian elimination method, and the complexityof each facet-defining inequality polynomially bounded by log a1, · · · , log an−1, and log b. Hence,the lemma is proved.
Let Q2 := {x ∈ Zn−1+ :
∑n−1i=1 aixi = b}. Lemma 27 and Lemma 28 show that P ′
2 = ∅ if andonly if Q2 6= ∅, which implies a polynomial reduction from the knapsack problem to Problem 25.
Lemma 27. If Q2 6= ∅, then P ′2 = ∅.
Proof. Let (w1, · · · , wn−1) ∈ Q2. Then∑n−1
i=1 aiwi = b and wi ≥ 0 for i = 1, · · · , n − 1. Letd := (w1, · · · , wn−1, 1) ∈ Zn. So dvk = wk
2rB + 12r −
b2rBA for k = 1, · · · , n− 1. Since 0 < B − b
A ≤ 1
by the definition of B, we get that 0 < 12r − b
2rBA ≤ 12rB . This implies 0 < dvk ≤ wk+1
2rB . Becausewk ≤ akwk ≤ b, we have wk + 1 ≤ 2b. Hence, 0 < dvk < 1 for k = 1, · · · , n − 1. Besides, dvn = 1
2and dvn+1 = 1. We now know that dv1, · · · , dvn+1 are all positive, so dx ≥ 1 is valid for P ′
2. Inaddition, xn ≤ rb+1 = 2016b2 + 3
2 is valid for P2, so xn ≤ rb+ 12 = 2016b2 +1 is valid for P ′
2. SinceP2 ∩ {x ∈ Rn : dx ≥ 1} = {vn+1} and the last component of vn+1 is greater than rb+ 1
2 , we knowP2 ∩ {x ∈ Rn : dx ≥ 1, xn ≤ rb+ 1
2} = ∅. Therefore P ′2 = ∅.
Lemma 28. If P ′2 = ∅, then Q2 6= ∅.
Proof. Let u1 := 1A
∑n−1i=1 aiv
i. Then u1 = ( a12rBA , · · · ,
an−1
2rBA ,12r − b
2rBA) ∈ P2. Let u2 := 12v
n +12v
n+1 = (0, · · · , 0, 34) and u3 := 13u
1 + 23u
2 = ( a16rBA , · · · ,
an−1
6rBA ,16r +
12 −
b6rBA ). Thus, both u2 and
u3 are in P2. If P ′2 = ∅, at least one Chvatal inequality is violated by u3. In other words, there
exists an inequality dx ≤ d0 + α valid for P2 such that (d, d0) ∈ Zn+1, 0 < α < 1, and d0 < du3.We claim that d and d0 satisfy the following four properties:
1)∑n−1
i=1 aidi = bdn.
2) dn = −1.
3) d0 = −1.
15
4) di ≤ 0 for i = 1, · · · , n− 1.
Then, (−d1, · · · ,−dn−1) ∈ Q2, so Q2 6= ∅.Let ∆ :=
∑n−1i=1 aidi − bdn. Then ∆ is an integer. Note that r∆+ 1
2dn − 1 < ⌊dvn⌋ ≤ d0. Thenr∆+ 1
2dn − 1 < d0 < du3 = 16rBA∆+ (12 + 1
6r )dn, and thus
6r(
r − 16rBA
)
∆− 6r < dn. (10)
Besides, −r∆+ dn − 1 < ⌊dvn+1⌋ ≤ d0. Since −r∆+ dn − 1 < d0 < du3 = 16rBA∆+ (12 +
16r )dn, we
get(
12 −
16r
)
dn < 1 +(
r + 16rBA
)
∆. (11)
Suppose that ∆ 6= 0 for the sake of contradiction. There are four cases to consider; ∆ > 0 anddn ≥ 0; ∆ > 0 and dn < 0; ∆ < 0 and dn ≥ 0; and ∆ < 0 and dn < 0.
Case 1: ∆ > 0 and dn ≥ 0.We know (12 − 1
6r ) >13 . By (11), we get dn < (3r + 1
2rBA)∆ + 3. Together with (10), we have
6∆r2 − (3∆ + 6)r − 1BA∆− 1
2rBA∆ < 3.
Note that −6∆r ≤ −6r and −2∆ ≤ − 1BA∆− 1
2rBA∆. Hence, we obtain the following:
∆(6r2 − 9r − 2) < 3,
which cannot be true, because r ≥ 2016 and ∆ ≥ 1. Therefore, Case 1 is not possible.
Case 2: ∆ > 0 and dn < 0. By (10), 6∆r2− 1BA∆− 6r < dn. Since ∆ ≥ 1, −∆r2 ≤ − 1
BA∆ andwe get 5∆r2 − 6r < dn. Because ∆ ≥ 1 and r ≥ 2016, dn > 0, which contradicts the assumptiondn < 0.
Case 3: ∆ < 0 and dn ≥ 0. Since ∆ ≤ −1 and 16rBA > 0, the right hand side of the inequality
(11) is less than 1 − r which has negative value. Because 12 > 1
6r , the inequality (11) implies thatdn be also negative, which contradicts the assumption dn ≥ 0.
Case 4: ∆ < 0 and dn < 0. Notice that 12rBA∆+ 1
2rdn−1 < ⌊du1⌋ ≤ d0. Then1
2rBA∆+ 12rdn−1 <
d0 < du3 = 16rBA∆+ (12 + 1
6r )dn. Then1
3rBA∆− 1 < (12 − 13r )dn < 1
3dn, and thus 1rBA∆− 3 < dn.
By (11) and the assumption dn < 0, we also know that dn < 1 + (r + 16rBA)∆. Then, we obtain
−4 <(
r − 56rBA
)
∆.
Since ∆ ≤ −1, the right hand side of the above inequality is at most −r+ 56rBA , which is less than
1− r. Then we get −4 < 1− r, but this cannot be true because r ≥ 2016.
All the four cases are not possible, so we get contradiction to the supposition that ∆ 6= 0.Therefore, ∆ = 0, i.e., (d, d0) satisfies the first property. In this case, du1 = 1
2rdn, du3 = (12+
16r )dn,
dvn = 12dn, and dvn+1 = dn. If dn = 0, then d0 satisfies d0 < du3 = 0 < d0 + 1 which is not true
since d0 is an integer. Hence, dn 6= 0. If dn ≥ 1, then the following relation holds.
⌊du3⌋ = ⌊(12 + 16r )dn⌋ < dn = ⌊dn⌋ = ⌊dvn+1⌋ ≤ ⌊d0 + α⌋ = d0.
However, we assumed that d0 < du3, and this implies d0 ≤ ⌊du3⌋. We got contradiction, sodn ≤ −1. Note that 1
2rdn − 1 < ⌊du1⌋ ≤ d0. Since d0 < du3 = (12 + 16r )dn, we get −1 < (12 − 1
3r )dn
16
and thus −2 ≤ dn. If dn = −2, ⌊dvn⌋ = −1 and ⌊du3⌋ = −2. Then ⌊du3⌋ < ⌊dvn⌋ ≤ d0, butthis contradicts the observation d0 ≤ ⌊du3⌋. Therefore, dn = −1, i.e., (d, d0) satisfies the secondproperty.
Since dn = −1, du3 = −12 − 1
6r . This means −1 < du3 < 0, so it also implies d0 = −1 whichis the third property. To prove the fourth property, let us consider dvk for k = 1, · · · , n − 1.dvk = dk
2rB − ( 12r −
b2rBA ) < d0 + 1 = 0. Then, dk < B − b
A . Since B is the smallest integer greater
than bA , B ≤ b
A + 1. Therefore, dk < 1 and thus dk ≤ 0 for k = 1, · · · , n− 1.
Theorem 29. Deciding emptiness of the Chvatal closure of a rational simplex is NP-complete.
Proof. The decision problem is obviously in NP. The NP-completeness is implied by Lemma 26,Lemma 27 and Lemma 28.
4.3 Optimization and separation over Chvatal closure
Eisenbrand [16] showed that the following separation problem is NP-hard, answering an earlyquestion of Schrijver [31].
Separation problem over the Chvatal closure. Given a rational polyhedron P ⊂ Rn and apoint x ∈ Qn, either show that x ∈ P ′ or find a valid Chvatal inequality dx ≤ d0 for P ′ such thatdx > d0.
According to a general result given by Grotschel, Lovasz and Schrijver [20], the above separationproblem and the following optimization problems are equivalent up to a polynomial time overhead.
Optimization problem over the Chvatal closure. Given a rational polyhedron P ⊂ Rn andan objective c ∈ Qn, find a vector x∗ ∈ P ′ satisfying cx∗ = max{cx : x ∈ P ′}, or show P ′ = ∅, orfind a ray z of the recession cone of P ′ for which cz is positive.
Theorem 24 and Theorem 29 trivially imply the NP-hardness of the optimization problem overthe Chvatal closure of a rational polytope in the unit hypercube or over the Chvatal closure of arational simplex, so we have the following conclusion.
Corollary 30. The optimization and separation problems over the Chvatal closure of a rationalpolytope in the unit hypercube or over the Chvatal closure of a rational simplex are NP-hard.
5 Deciding whether Chvatal closure equals integer hull
A result of Cornuejols and Li [9] is that it is NP-hard to decide whether the Chvatal rank of arational polyhedron is 1. In this section, we improve their result by showing that the decisionproblem remains NP-hard even when the input polyhedron is contained in the unit hypercube oris a rational simplex.
Theorem 31. Let P ⊆ [0, 1]n be a rational polytope. It is NP-hard to decide whether P ′ = PI .
Proof. Recall that P1 contains no integer point by Lemma 21. That means (P1)I = ∅. By Lemma 22and Lemma 23, P ′
1 = (P1)I if and only if there exists a solution to {x ∈ Zn−4+ :
∑n−4i=1 aixi =
b}. Therefore, there exists a polynomial reduction from the knapsack problem to the problem ofdeciding whether a given rational polytope P ⊆ [0, 1]n satisfies P ′ = PI , so the latter problem isalso NP-hard.
17
Now, we turn our attention to simplices.
Theorem 32. Let P ⊂ Rn be a rational simplex. It is NP-hard to decide whether P ′ = PI .
Proof. Recall that the polyhedron P2 constructed in Section 4.2 is a rational simplex, and Lemma 27and Lemma 28 show that P ′
2 is empty if and only if there exists a solution to {x ∈ Zn+ :
∑n−1i=1 aixi =
b}, thereby providing a polynomial reduction from the knapsack problem to the problem of decidingwhether P ′ = ∅ for a given rational simplex P .
To prove Theorem 32, it suffices to show that P2 contains no integer point, which means thatP ′2 = ∅ is equivalent to P ′
2 = (P2)I . Suppose that P2 contains an integer point for the sake ofcontradiction. Then, there exists an integer d ∈ [−rb+ 1
2 , rb+1] such that P2(d) := {x ∈ P2 : xn =d} contains an integer point.
First of all, we assume that d > 0. The hyperplane defined by xn = d intersects the line segmentsbetween vn+1 and vi for i = 1, · · · , n. In fact, P2(d) is the convex hull of those n intersection points.Let ui0 denote the intersection point of {x ∈ Rn : xn = d} and the line segment between vn+1 andvi for i = 1, · · · , n. Then, for i = 1, 2, · · · , n− 1,
ui0 :=−rd+ 1
2− b
2BA
rb+1− 1
2r+ b
2rBA
(a1, · · · , an−1, 0) +
(
12rB −
d− 1
2r+ b
2rBA
2rB(rb+1− 1
2r+ b
2rBA)
)
ei +den, and
un0 :=−rd+ 3
4r
rb+ 1
4
(a1, · · · , an−1, 0) +den.
Let a and ui, where i = 1, · · · , n, denote the projection of (a1, · · · , an−1, 0) and ui0 to the spacedefined by the first n− 1 components, respectively. Then, a = (a1, · · · , an−1) ∈ Zn−1 and
ui :=−rd+ 1
2− b
2BA
rb+1− 1
2r+ b
2rBA
a +
(
12rB −
d− 1
2r+ b
2rBA
2rB(rb+1− 1
2r+ b
2rBA)
)
ei for i = 1, 2, · · · , n − 1,
un :=−rd+ 3
4r
rb+ 1
4
a.
Since P2(d) contains an integer point with the last component equal to d, the convex hull ofu1, · · · , un, which is a simplex in Rn−1 and is denoted by S, contains an integer point in Zn−1.Consider the line L′ through 0 and a in Rn−1. By a simple calculation, we see that the intersectionof L′ and S is the line segment between un and 1
A
∑n−1i=1 aiu
i. Let u denote 1A
∑n−1i=1 aiu
i. Then
u =
(
−rd+ 1
2− b
2BA
rb+1− 1
2r+ b
2rBA
+ 12rBA −
d− 1
2r+ b
2rBA
2rBA(rb+1− 1
2r+ b
2rBA)
)
a.
Now we show that the line segment between un and u does not contain an integer point. Sincea1, · · · , an−1 are relatively prime integers, there is no integer point between ℓa and (ℓ + 1)a for
any ℓ ∈ Z. Let C1 :=−rd+ 3
4r
rb+ 1
4
and C2 :=
(
−rd+ 1
2− b
2BA
rb+1− 1
2r+ b
2rBA
+ 12rBA −
d− 1
2r+ b
2rBA
2rBA(rb+1− 1
2r+ b
2rBA)
)
. Then
un = C1a and u = C2a. We will show that there exists an integer ℓ such that C1, C2 ∈ (ℓ, ℓ + 1).By doing so, we can prove that there is no integer point in the line segment between un and u.
Note that d can be expressed as kb + h for some 0 ≤ k ≤ 2016b and 0 ≤ h < b. Then we canrewrite both un and u as follows:
un =(
−k +−rh+ 1
4k+ 3
4r
rb+ 1
4
)
a =(
−k − 1 +r(b−h)+ 1
4+ 1
4k+ 3
4r
rb+ 1
4
)
a,
18
u =
(
−k +−rh+(r−k)( 1
2r− b
2rBA)+k
rb+1− 1
2r+ b
2rBA
+ 12rBA −
d− 1
2r+ b
2rBA
2rBA(rb+1− 1
2r+ b
2rBA)
)
a
=
(
−k − 1 +r(b−h)+(r−k−1)( 1
2r− b
2rBA)+(k+1)
rb+1− 1
2r+ b
2rBA
+ 12rBA −
d− 1
2r+ b
2rBA
2rBA(rb+1− 1
2r+ b
2rBA)
)
a.
In the following, we consider three possible cases: h = 0; h = 1 and k = 2016b = r− 12b ; and h ≥ 1
and k ≤ 2016b − 1 = r − 1− 12b .
Case 1: h = 0. In this case, the integer part of C1 is −k, while its fractional part is1
4k+ 3
4r
rb+ 1
4
since it
is certainly positive and less than 1. Notice that 12r −
b2rBA =
B− bA
2rB ≤ 12rB , because B is the smallest
integer greater than bA . Then (r−k)( 1
2r−b
2rBA)+k ≤ 12B+r. In addition, 0 < 1−
d− 1
2r+ b
2rBA
rb+1− 1
2r+ b
2rBA
< 1,
because 0 < d < rb+ 1. Therefore,−rh+(r−k)( 1
2r− b
2rBA)+k
rb+1− 1
2r+ b
2rBA
+ 12rBA −
d− 1
2r+ b
2rBA
2rBA(rb+1− 1
2r+ b
2rBA)is positive
and at most 12rB + 1
2rBA which is less than 1. That means the integer part of C2 is −k and itsfractional part is positive. In this case, C1, C2 ∈ (−k,−k + 1).
Case 2: h = 1 and k = 2016b = r− 12b . Since k < r, 0 < r(b− 1) + 1
4 +14k+
34r < rb+ 1
4 . Thus,
we get 0 <r(b−1)+ 1
4+ 1
4k+ 3
4r
rb+ 1
4
< 1. Then C1 ∈ (−k − 1,−k). Note that
r(b− h) + (r − k − 1)( 12r − b
2rBA) + (k + 1) = r(b− 1) + ( 12b − 1)( 1
2r − b2rBA) + 2016b + 1
= rb+ 1− 12r +
b2rBA + 1
2b (−1 + 12r −
b2rBA).
In addition,1
2rBA −d− 1
2r+ b
2rBA
2rBA(rb+1− 1
2r+ b
2rBA)
= 14rBA(rb+1− 1
2r+ b
2rBA).
In this case,
r(b−h)+(r−k−1)( 1
2r− b
2rBA)+(k+1)
rb+1− 1
2r+ b
2rBA
+ 12rBA −
d− 1
2r+ b
2rBA
2rBA(rb+1− 1
2r+ b
2rBA)
= 1 +1
2b(−1+ 1
2r− b
2rBA)+ 1
4rBA
rb+1− 1
2r+ b
2rBA
.
is less than 1, because 12r−
b2rBA+
14rBA ≤ 1
2rB+ 14rBA < 1
2b . Therefore, we get that C2 ∈ (−k−1,−k).
Case 3: h ≥ 1 and k ≤ 2016b − 1 = r − 1 − 12b . As in the previous case, we can show that
C1 ∈ (−k − 1,−k). Notice that
r(b− h) + (r − k − 1)( 12r − b
2rBA ) + (k + 1) ≤ rb− 12b +
12b(
12r −
b2rBA )
= rb+ 1− 12r +
b2rBA − (1 + 1
2b )(1−12r +
b2rBA).
We also have the following:
12rBA −
d− 1
2r+ b
2rBA
2rBA(rb+1− 1
2r+ b
2rBA)
≤ 12rBA ≤ 1
2rb ≤ 1rb+1− 1
2r− b
2rBA
.
Since 1− (1 + 12b)(1 −
12r +
b2rBA) < 0, we get
r(b−h)+(r−k−1)( 1
2r− b
2rBA)+(k+1)
rb+1− 1
2r+ b
2rBA
+ 12rBA −
d− 1
2r+ b
2rBA
2rBA(rb+1− 1
2r+ b
2rBA)
< 1.
It is obvious that r(b− h) + (r − k − 1)( 12r −
b2rBA) + (k + 1) > 0, so C2 ∈ (−k − 1,−k).
19
We have proved that there is no integer point in the line segment connecting un and u. Thenany integer point in conv{u1, · · · , un} is not on the line segment. Let v ∈ conv{u1, · · · , un}∩Zn−1.Observe that there exists δ ∈ R such that ‖v − δa‖∞ ≤ 1
rB , because ‖ui − u‖∞ ≤ 1rB for i =
1, · · · , n−1, ‖un−un‖∞ = 0, and v ∈ conv{u1, · · · , un}. In addition, v is not on the line connecting0 and a in Rn−1. That means there exists an index j ∈ {1, · · · , n−2} such that (vj , vn−1) is not onthe line through 0 and (aj , an−1) in R2 [9]. The following can be proved with a simple geometricanalysis in R2:
Claim 1. Let u := (u1, u2) 6= (0, 0) and v := (v1, v2) be two points in Z2. Then, for any δ ∈ Rn,
‖v − δu‖∞ ≥ |u1v2−u2v1||u1|+|u2|
.
By Claim 1,
‖(vj , vn−1)− δ(aj , an−1)‖∞ ≥|aj vn−1−an−1vj |
|aj |+|an−1|.
for any δ ∈ R. Because (vj , vn−1) is not on the line through 0 and (aj , an−1), |aj vn−1−an−1vj | ≥ 1.Since 0 < aj < an−1 ≤ b,
‖(vj , vn−1)− δ(aj , an−1)‖∞ > 12b . (12)
Since ‖v− δa‖∞ is at least as large as ‖(vj , vn−1)− δ(aj , an−1)‖∞, we now know that ‖v− δa‖∞ isgreater than 1
2b by (12) for all δ ∈ R. However, this contradicts the observation that ‖v−δa‖∞ ≤ 1rB
for some δ ∈ R. Therefore, we conclude that P2(d) does not contain an integer point when d > 0.Now, we assume that d ≤ 0. Then the hyperplane {x ∈ Rn : xn = d} intersects the line
segment between vn and vi for i = 1, · · · , n − 1, n + 1. We denote each intersection point by wi0
accordingly. Then we can calculate, for 1 ≤ i ≤ n− 1,
wi0 :=
rd− 1
2+ b
2BA
−rb+ 1
2− 1
2r+ b
2rBA
(a1, · · · , an−1, 0) +
(
12rB −
d− 1
2r+ b
2rBA
2rB(−rb+ 1
2− 1
2r+ b
2rBA)
)
ei +den , and
wn+10 :=
−rd+ 3
4r
rb+ 1
4
(a1, · · · , an−1, 0) +den.
We know that P2(d) is the convex hull of the above n intersection points. Let wi, where i =1, · · · , n− 1, n+1, denote the projection of wi
0 to the space defined by the first n− 1 components,respectively. Then,
wi :=rd− 1
2+ b
2BA
−rb+ 1
2− 1
2r+ b
2rBA
a +
(
12rB −
d− 1
2r+ b
2rBA
2rB(−rb+ 1
2− 1
2r+ b
2rBA)
)
ei for 1 ≤ i ≤ n− 1,
wn+1 :=−rd+ 3
4r
rb+ 1
4
a.
Since P2(d) contains an integer point, its projection, which is identical to the convex hull ofw1, · · · , wn−1, wn+1, contains an integer point in Rn−1. Let w denote 1
A
∑n−1i=1 aiw
i. Then w canbe written as
w =
(
rd− 1
2+ b
2BA
−rb+ 1
2− 1
2r+ b
2rBA
+ 12rBA −
d− 1
2r+ b
2rBA
2rBA(−rb+ 1
2− 1
2r+ b
2rBA)
)
a.
In fact, the line though 0 and a in Rn−1 intersects with conv{w1, · · · , wn−1, wn+1} in the linesegment between wn+1 and w. As in the previous case, we will show that the line segment between
wn+1 and w does not contain an integer point. Let D1 :=−rd+ 3
4r
rb+ 1
4
and D2 :=rd− 1
2+ b
2BA
−rb+ 1
2− 1
2r+ b
2rBA
+ 12rBA
−d− 1
2r+ b
2rBA
2rBA(−rb+ 1
2− 1
2r+ b
2rBA). We will prove that both D1,D2 ∈ (ℓ, ℓ+1) for some integer ℓ. Since d can
20
be written as kb+ h for some −2016b ≤ k ≤ −1 and 0 ≤ h < b, we can rewrite both wn+1 and was follows:
wn+1 =(
−k +−rh+ 1
4k+ 3
4r
rb+ 1
4
)
a =(
−k − 1 +r(b−h)+ 1
4+ 1
4k+ 3
4r
rb+ 1
4
)
a,
w =
(
−k +rh+ 1
2k−(r+k)( 1
2r− b
2rBA)
−rb+ 1
2− 1
2r+ b
2rBA
+ 12rBA −
d− 1
2r+ b
2rBA
2rBA(rb+1− 1
2r+ b
2rBA)
)
a
=
(
−k − 1 +r(b−h)− 1
2−(r+k−1)( 1
2r− b
2rBA)+ 1
2k
rb− 1
2+ 1
2r− b
2rBA
+ 12rBA −
d− 1
2r+ b
2rBA
2rBA(rb+1− 1
2r+ b
2rBA)
)
a.
There are two cases to consider: h = 0; h > 0. When h = 0, it is straightforward to show thatD1,D2 ∈ (−k,−k+1). When h > 0, it is also easy to show that D1,D2 ∈ (−k− 1,−k). Therefore,the line segment between wn+1 and w does not contain any integer point. The rest of the proof issimilar to that given in the previous case when d is positive.
6 Deciding whether adding a certain number of Chvatal cuts can
yield integer hull
Let k be any given positive integer. In this section, we prove that it is NP-hard to decide whetheradding at most k Chvatal inequalities to the description of a rational polytope suffices to describeits integer hull. In particular, we show that this is true for the rational polytopes in the unithypercube and for the rational simplices.
6.1 The case of polytopes contained in the unit hypercube
Theorem 33. Let P ⊂ [0, 1]n be a rational polytope and k be a positive integer. Deciding if we canobtain PI by adding at most k Chvatal inequalities to the linear description of P is NP-hard.
To prove Theorem 33, we use Lemma 34 for the case k = 1 and Lemma 35 for k ≥ 2.
Lemma 34. Let P ⊆ [0, 1]n be a rational polytope. Deciding whether there exists a Chvatal in-equality of P such that adding it to the linear description of P gives PI is NP-complete.
Proof. We consider the partition problem:
Partition Problem (see [17]). Given positive integers a1, · · · , an, is there a subset K of the setof indices [n] such that
∑
i∈K ai =∑
j∈[n]\K aj?
The partition problem is NP-complete. Mahajan and Ralphs [27] constructed a rational polytopein the unit hypercube using the input from an instance of the partition problem. We borrow theirconstruction for the proof. Let a1, · · · , an−2 be the input for an instance of the partition problemwith index set [n − 2]. Let ak denote 1∑n−2
j=1ajak for k = 1, · · · , n − 2. We use the following n + 2
vectors in [0, 1]n:
21
v1 := ( 12 +
12(n−1) ,
12(n−1) , · · · , 1
2(n−1) , 0, 0 )
v2 := ( 12(n−1) ,
12 +
12(n−1) , · · · , 1
2(n−1) , 0, 0 )...
vn−2 := ( 12(n−1) ,
12(n−1) , · · · , 1
2 +1
2(n−1) , 0, 0 )
vn−1 := ( a1, a2, · · · , an−2, 1, 1 )vn := ( a1, a2, · · · , an−2,
12 −
12∑n−2
j=1aj, 0 )
vn+1 := ( a1, a2, · · · , an−2, 0, 12 − 1
2∑n−2
j=1aj
)
vn+2 := ( 0, 0, · · · , 0, 12 , 0 )
Let P3 := conv{v1, · · · , vn+2} be the convex hull of the above n + 2 vectors. Mahajan andRalphs [27] proved that there is (π, π0) ∈ Zn such that P3 ⊆ {x ∈ Rn : π0 < πx < π0 + 1} if andonly if there exists a subset K of [n − 2] such that
∑
i∈K ai =∑
j∈[n−2]\K aj, thereby showing thedecision problem for the existence of such (π, π0) is NP-complete.
To complete the proof, we show that P3 contains no integer point. Let d := (1, · · · , 1, 1,−1).Then dvi = 1 − 1
2(n−1) for i = 1, · · · , n − 2. Besides, we can easily calculate dvn−1 = 1, dvn =32 − 1
2∑n−2
j=1aj, dvn+1 = 1
2 + 12∑n−2
j=1aj, and dvn+2 = 1
2 . Without loss of generality, we assume∑n−2
j=1 aj ≥ 1. So 0 < dx < 2 is valid for all x ∈ P3, and thus dx = 1 is valid for P ′3.
Since 0 < a1 <∑n−2
j=1 aj by the assumption, 0 < a1 < 1. This implies that the first component
of each vi be less than 1, so x1 ≤ 0 is valid for P ′3. Notice that P3∩{x ∈ [0, 1]n : x1 ≤ 0} = {vn+2}.
Besides, dvn+2 = 12 6= 1. Since P ′
3 ⊆ P3 ∩ {x ∈ [0, 1]n : dx = 1, x1 ≤ 0} and P3 ∩ {x ∈ [0, 1]n :dx = 1, x1 ≤ 0} = ∅, we see P ′
3 = ∅. Hence, (P3)I = ∅.
Lemma 35. Let P ⊆ [0, 1]n be a rational polytope and k be a positive integer. Deciding if we canobtain PI by adding at most k Chvatal inequalities of P to the linear description of P is NP-hard.
Proof. Let a1, · · · , an−4, b be the input for an instance of the knapsack problem of dimension n−4.Recall that, in Section 4.1, (w1, · · · , wn−4) ∈ Q1 = {x ∈ Zn−4
+ :∑n−4
j=1 ajxj = b} if and onlyif P ′
1 is empty. When P ′1 is empty, we observe that two Chvatal inequalities dx ≥ 1 where d =
(w1, · · · , wn−4,−∑n−4
i=1 wi, 1,−1, 1) and xn−3 + xn−2 + xn−1 + xn ≤ 3 remove every point in P1.Then, we can claim that the knapsack instance Q1 has a solution if and only if adding at most kChvatal inequalities is sufficient to describe the integer hull of P1 which is empty for any k ≥ 2.Therefore, the decision problem is NP-complete.
Theorem 33 follows from Lemma 34 and Lemma 35. Note from the proof of Lemma 35 thatthe lemma also holds in the case that k is not a constant.
6.2 The case of simplices
Theorem 36. Let P ⊂ Rn be a rational simplex and k be a positive integer. Deciding if we canobtain PI by adding at most k Chvatal inequalities to the linear description of P is NP-hard.
To prove Theorem 36, we use Lemma 37 for the case k = 1 and Lemma 38 for k ≥ 2.
22
Lemma 37. Let P ⊆ [0, 1]n be a rational polytope. Deciding whether there exists a Chvatal in-equality of P such that adding it to the linear description of P yields PI is NP-complete.
Proof. Mahajan and Ralphs [27] constructed a simplex in Rn using the input from an instance ofthe partition problem. Again, we borrow their construction for the proof. Let a1, · · · , an−1 be theinput for an instance of the partition problem with index set [n−1]. Let n+1 vectors v1, · · · , vn+1
be defined as follows:
v1 := ( 12 + 1
2n ,12n , · · · , 1
2n ,12n )
v2 := ( 12n ,
12 + 1
2n , · · · , 12n ,
12n )
...vn−1 := ( 1
2n ,12n , · · · , 1
2 + 12n ,
12n )
vn := ( 12n ,
12n , · · · , 1
2n ,12 +
12n )
vn+1 := ( a1, a2, · · · , an−1,12 −
12
∑n−1i=1 ai )
Let P4 := conv{v1, · · · , vn+1} be the convex hull of these n+ 1 vectors. Mahajan and Ralphs [27]proved that there is (π, π0) ∈ Zn+1 such that P4 ⊆ {x ∈ Rn : π0 < πx < π0 + 1} if and only ifthere exists a subset K of [n− 1] such that
∑
i∈K ai = s.As in the proof of Lemma 34, it suffices to show that the integer hull of P4 is empty. Suppose
that P4 ∩ Zn 6= ∅ for the sake of contradiction. Let d := (1, · · · , 1, 2). Notice that dvn+1 = 1,dvn = 3
2 + 12n , and dvi = 1 + 1
2n for i = 1, · · · , n − 1. Then 1 < dvi < 2 for i = 1, · · · , n. Because1 ≤ dx < 2 for all x ∈ P4, dz = 1 for all z ∈ P4 ∩ Zn. Let z∗ ∈ P4 ∩ Zn. Since z∗ ∈ P4, z
∗ is aconvex combination of v1, v2, · · · , vn+1 with nonnegative multipliers α1, α2, · · · , αn+1, respectively.Note that dz = 1 implies α1 = · · · = αn = 0 and thus z = vn+1. However, vn+1 is not integral,which contradicts the integrality of z∗. Hence, (P4)I = ∅.
Lemma 38. Let P ⊆ Rn be a rational simplex and k be a number greater than 1. Deciding ifwe can obtain PI by adding at most k Chvatal inequalities of P to the linear description of P isNP-hard.
Proof. Let a1, · · · , an−1, b be the input from an instance of the knapsack problem of dimensionn − 1. Recall that, in Section 4.2, (w1, · · · , wn−1) ∈ Q2 = {x ∈ Zn−1
+ :∑n−1
j=1 ajxj = b} if andonly if P ′
2 is empty. Also, when P ′2 is empty, adding two Chvatal inequalities dx ≥ 1, where
d = (w1, · · · , wn−1, 1), and xn ≤ rb + 12 to the description of P2 results in an empty polytope.
Then, we can claim that the knapsack instance Q2 is feasible if and only if adding at most kChvatal inequalities is sufficient to describe the integer hull of P2 for any k ≥ 2. Therefore, theNP-completeness is proved.
Theorem 36 follows from Lemma 37 and Lemma 38. We also know from the proof of Lemma 38that the lemma is true even when k is not a constant.
Acknowledgments: This work was supported in part by NSF grant CMMI 1560828 and ONRgrant 00014-15-12082.
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